Derivative of Inverse Trig Function with Square Root in Denominator

[ttttrigg3r]

Homework Statement

tan^-1(x/(1-x^2)^1/2) find the derivative

the problem comes from 3g from MIT's PDF I found
http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/part-b-implicit-differentiation-and-inverse-functions/problem-set-2/MIT18_01SC_pset5sol.pdf

here is the solution key
http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/part-b-implicit-differentiation-and-inverse-functions/problem-set-2/MIT18_01SC_pset5prb.pdf

Homework Equations

The solution key says if y = x/(1-x^2)^1/2 then y' = (1-x^2)^-3/2 When I do it, y' comes out differently. This is how I attempt to solve it.

The Attempt at a Solution

Original problem: tan^-1((x/(1-x^2)^1/2)) what is the derivative with respect to x?

let y=x/(1-x^2)^1/2 y=x*(1-x^2)^-1/2 using the product rule I get:
y'=(1)(1-x^2)^-1/2 + x(-1/2)((1-x^2)^-3/2)(-2)
y'=(1-x^2)^-1/2 + (x^2)(1-x^2)^-3/2
This looks a lot different than the y' stated in the solution manual: y' = (1-x^2)^-3/2

Am I on the right track? Is the solution manual wrong? or am I missing a step? Thank you ahead.

 

[hunt_mat]

You need to use the chain rule, you have:
$$f(x)=\tan^{-1}\left[ \frac{x}{\sqrt{1-x^{2}}} \right]$$
write (as you did) y=x/(1-x^2)^1/2 and use the chain rule:
$$\frac{df}{dx}=\frac{df}{dy}\frac{dy}{dx}$$
I believe you have already computed the second factor.

 

[ttttrigg3r]

Thank you for the reply. However my question is why is my computed derivative of the inside term different than the answer key?

 

[hunt_mat]

It's not different at all, take out a factor of $1/\sqrt{1-x^{2}}$ and you will see (with a little work) why they are the same.

 

[ttttrigg3r]

Oh nice. I was wondering what I could do. Ty.